The Exponent Rule for Derivatives — Your One-Stop Shop to Staircase-Looking Functions

By mathvault | Calculus

Math Vault — The Exponent Rule for Derivative

Like the featured image above? Good for you! As students (fanatics?) of mathematics and other technical sciences, we have an intuitive appreciation of the role this topic — calculus — plays out in our daily life. While obviously not a subject to be mastered overnight, what’s true is that our first exposure to it tends to be almost universally restricted to the topic of derivatives of elementary functions

Interestingly enough,  when it comes to the calculation of derivatives, there is a rule of thumb out there that goes something like this: either the function is basic, in which case we can appeal to the table of derivatives, or the function is composite, in which case we can differentiated it recursively — by breaking it down into the derivatives of its constituents via a series of derivative rules.

Indeed, for those of us who have ever read a calculus textbook or attended a calculus course, we might be already familiar with most — if not all — of these derivative rules:

If the function $cf$ is defined on an interval $I$ and $f$ is differentiable on $I$, then $\displaystyle (cf)’=cf’$ on $I$.

Given a real number $r$ greater or equal to $1$ , $(x^r)’ = r x^{r-1}$ for all $x \in \mathbb{R}$. In the case where $r$ is less than $1$ (and non-zero), $(x^r)’ = r x^{r-1}$ for all $x \ne 0$.

If the function $f+g$ is well-defined on an interval $I$, with $f$ and $g$ being both differentiable on $I$, then $\displaystyle (f+g)’ = f’ + g’$ on $I$.

If the function $f-g$ is well-defined on an interval $I$, with $f$ and $g$ being both differentiable on $I$, then $\displaystyle (f-g)’ = f’ – g’$ on $I$.

If the function $fg$ is well-defined on an interval $I$, with $f$ and $g$ being both differentiable on $I$, then $\displaystyle (fg)’ = f’g + fg’$ on $I$.

If the function $\displaystyle \frac{1}{f}$ is well-defined on an interval $I$, with $f$ being differentiable on $I$, then $\displaystyle \left( \frac{1}{f} \right)’ = \frac{-f’}{f^2}$ on $I$.

If the function $\displaystyle \frac{f}{g}$ is well-defined on an interval $I$, with $f$ and $g$ being both differentiable on $I$, then $\displaystyle \left( \frac{f}{g} \right)’ = \frac{f’g – fg’}{g^2}$ on $I$.

If the function $f \circ g$ in well-defined on an interval $I$, with $g$ being differentiable on $I$ and $f$ being differentiable on $g(I)$, then:

\begin{align*} [(f \circ g)(x)]’ = f'[g(x)] \, g'(x) \qquad (x \in I) \end{align*}

(refer to the module on Chain Rule for more detail)

If the function $f^{-1}$ is well-defined on an interval $I$, and the expression $\displaystyle \frac{1}{ f'[ f^{-1} (x) ] }$ makes sense on $I$ (i.e., for all $x \in I$, $f$ is differentiable at $f^{-1} (x)$, with the derivative being non-zero), then:

\begin{align*} [f^{-1}(x)]’ = \frac{1}{f'[f^{-1}(x)]} \qquad (x \in I) \end{align*}

(refer to the module on Inverse Function Theorem for more detail)

For the most part, these rules are more than enough to handle the vast majority of functions one would encounter. However, when we look at our repertoire of functions, we see that something else is missing, namely:

What about the functions constructed via exponentiation?

Here, prompted by an unusual sense of urgency, we proceed to toy around with the idea of derivative of a power, subsequently ending up developing a rule precisely for that purpose. The name? Exponent Rule of course! Now, we bet that you haven’t seen this one yet — at least not in a standard calculus textbook or Khan Academy anyway!

Table of Contents


I. Exponent Rule for Derivative — Theory
II. Exponent Rule for Derivative — Applications
1) $\pi^x$
2) Exponential Function (Arbitrary Base)
3) $ x^{\ln x}$
4) $(x^2+1)^{\sin x}$
5) $(2x)^{3x}$
6) ${\left( \sqrt{x}^{\cos x}\right)}^{\ln x}$
7) $\sqrt{x}^{\, \cos x^{\, \ln x}}$
III. Closing Words

Exponent Rule for Derivative — Theory

Given a base function $f$ and an exponent function $g$ — both defined on an interval $I$ — one can construct the power function $f^g$ (i.e., $f$ raised to $g$), which is defined as long as $f(x)>0$ on $I$ (why?). In which case, the properties of logarithm warrant that:

\begin{align*} f(x)^{g(x)} & = e^{\ln \left[ f(x)^{g(x)} \right] } = e^{g(x) \ln f(x)} \end{align*}

which suggests that $\displaystyle e^{g(x) \ln f(x)}$ and $\displaystyle f(x)^{g(x)}$ can be used interchangeably on $I$.

More importantly though, in the case where $f$ and $g$ are both differentiable on $I$, using a combination of Chain Rule and Product Rule, we get that:

\begin{align*} \left( f(x)^{g(x)} \right)’ & = \left( e^{g(x) \ln f(x)} \right)’ \\ & = e^{g(x) \ln f(x)} \, [g(x) \ln f(x)]’ \\ & = f(x)^{g(x)} \left[ g'(x) \ln f(x) + f'(x) \frac{ g(x)}{f(x)} \right] \end{align*}

Too much $f$ and $g$? No worry. All we have shown is that the derivative of $f^g$ can be expressed in terms of the derivatives of $f$ and $g$ — and hence the Exponent Rule for derivatives!

Theorem — The Exponent Rule for Derivative

Given a base function $f$ and an exponent function $g$, if:

  1. The power function $f^g$ is well-defined on an interval $I$ (i.e.,$f$ and $g$ both well-defined on $I$, with $f>0$ on $I$)
  2. Both $f$ and $g$ are differentiable on $I$

then the function $f^g$ is differentiable on $I$ as well. In addition:

\begin{align*} (f^g)’ = f^g \left( g’ \ln f + f’ \frac{g}{f} \right) \qquad (x \in I) \end{align*}

In case it’s still not clicking,  this is what the Exponent Rule reads in English:

The derivative of a power, is equal to the power itself times the following: the derivative of the exponent times the logarithm of the base, plus the derivative of the base times the exponent-base ratio..

Wonderful meditation mantra? Or maybe a bad-pun-turned-sour-joke? Either way, we just want you to know that there is definitely a pattern to this! Indeed, the Exponent Rule basically states that the derivative of $f^g$ takes the form of $f^g (A+B)$, where:

  • $A$ is obtained by taking the derivative of the exponent, times the logarithm of the base.
  • $B$ is obtained by taking the derivative of the base instead, times the ratio with the exponent on the top.

In fact, with just a bit of practice, it’s possible to master the Exponent Rule as much as we do with the Quotient Rule — and this is not to mention the whole new world it opens up to our mental calculus fanatics!

Exponent Rule for Derivative — Applications

Traditionally, to evaluate the derivative of a power function, one would have to either resort to logarithmic differentiation, or base-e standardisation before differentiating it. With the advent of the Exponent Rule, both of these approaches basically become obsolete — not because they are irrelevant per se, but because they have already been carried out during the derivation of the Exponent Rule.

In what follows, we lay out 7 examples illustrating how the Exponent Rule can be used to streamline the differentiation process for power functions, thereby freeing up some of our brainpower and resources to other potentially-more-challenging tasks that are yet to come.

Example 1 — $\pi^x$

With the base function being $\pi$ and the exponent function $x$, the exponential function $\pi^x$ — which is defined on the entire real line — easily satisfies the preconditions laid out by the Exponent Rule. In which case, differentiating it using the Rule yields that:

\begin{align*} (\pi^x)’ & = \pi^x \left[ (x)’ \ln \pi + (\pi)’ \frac{x}{\pi} \right] = \pi^x \ln \pi \qquad (x \in \mathbb{R})\end{align*}

which shows that $\pi^x$ differentiates to itself along with some adjustment factor — a typical behavior among the exponential functions.

Now, is that a heavy-handed differentiation approach to a particularly-simple function? Actually, not really, for as it turns out, the alternative approaches would have been essentially the same — save perhaps a few more extra “patching” steps.

Finally, to celebrate our first successful execution of the Exponent Rule, we’ve cooked up the graphs of $\pi^x$ and its derivative here:

The exponential function pi^x (red) and its derivative (green)

$\pi^x$ in red, and its derivative in green. Note that they are almost identical — up to a certain constant that is.

Example 2 — Exponential Function (Arbitrary Base)

Generalizing from the last example, we see that given the base function $b$ ($b>0, b \ne 1$) and the exponent function $x$, the exponential function $b^x$ — which is defined on the entire real line — can also be differentiated using the Exponent Rule as follows:

\begin{align*} (b^x)’ & = b^x \left[ (x)’ \ln b + (b)’ \frac{x}{b} \right] = b^x \ln b \qquad (x \in \mathbb{R})\end{align*}

which shows that in general, an exponential function differentiates to itself, along with the natural logarithm of its base.

Example 3 — $x^{\ln x}$

With the base function $x$ and the exponent function $\ln x$, the power function $\displaystyle x^{\ln x}$ unfortunately can be only defined on the set of positive real numbers $\mathbb{R_+}$. However, this is really more of a blessing than a curse, as $\mathbb{R_+}$ happens to be the biggest domain under which both the base and the exponent function are differentiable anyway. In which case, applying the Exponent Rule yields that:

\begin{align*} (x^{\ln x})’ & = x^{\ln x} \left[ (\ln x)’ (\ln x) + (x)’\frac{\ln x}{x} \right] \\ & = x^{\ln x} \left[ \frac{\ln x}{x} + \frac{\ln x}{x} \right] \\ & = x^{\ln x} \, \frac{2 \ln x}{x} \qquad (x>0) \end{align*}

which is not too bad as a result. To be sure, here’s the graph of $\displaystyle x^{\ln x}$ and its derivative alongside with it:

The Graphs of x^ln x (red) and its derivative

The graphs of $\displaystyle x^{\ln x}$ (red) with its derivative (green) alongside. Note that $\displaystyle x^{\ln x}$ attains the minimum at $(1,1)$.

Example 4 — $(x^2+1)^{\sin x}$

There we go. Time to unleash the extravaganza! With the base function $x^2+1$ and the exponent function $\sin x$, the power function $\displaystyle (x^2+1)^{\sin x}$ — which is defined everywhere on $\mathbb{R}$ — has the added bonus that both the base and the exponent functions are differentiable on $\mathbb{R}$ as well. In which case, a single application of the Exponent Rule yields:

\begin{align*} \left( (x^2+1)^{\sin x}\right)’ & = (x^2+1)^{\sin x} \left[ (\sin x)’ \ln (x^2+1) + (x^2+1)’ \frac{\sin x}{x^2+1} \right] \\ & = (x^2+1)^{\sin x} \left[ \cos x \ln (x^2+1) + \frac{2x \sin x}{x^2+1} \right] \end{align*}

Interesting! Let’s plug them into our graphing application and see what kind of madness it produces in return!

The graphs of the function (x^2+!)^{\sin x} and its derivative

Woah!! Looks like we’ve just created a pair of monsters! Nevertheless, rest assured that this is not a mindless mistake. As usual, the original function $\displaystyle (x^2+1)^{\sin x}$ is in red, and its derivative in green.

Example 5 — $(2x)^{3x}$

Running through our checklist as usual, we see that the base function $2x$ can only be positive on $\mathbb{R_+}$ (i.e., the set of positive real numbers). However, in doing so, we also found that both the base function $2x$ and the exponent function $3x$ are well-defined and differentiable on $\mathbb{R_+}$. As a result, the power function $\displaystyle (2x)^{3x}$ is differentiable on $\mathbb{R_+}$ as well, with its derivative obtainable through the use of Exponent Rule:

\begin{align*} \left[ (2x)^{3x}\right]’ & = (2x)^{3x} \left[ (3x)’ \ln (2x) + (2x)’ \frac{3x}{2x} \right] \\ & = (2x)^{3x} \left[ 3 \ln(2x) + 3 \right] \qquad (x>0) \end{align*}

Alternatively, we can also push the $3$ in the exponent function into the base function, as follows:

\begin{align*} (2x)^{3x} & = \left[{(2x)^3}\right]^x = (8x^3)^x \qquad (x>0)\end{align*}

in which case, the previous observations about the domain and differentiability still hold, and applying the Exponent Rule once again yields:

\begin{align*} \left[ (8x^3)^x \right]’ & = (8x^3)^x  \left[ \ln (8x^3) + 24x^2 \frac{x}{8x^3} \right] \\ & = (2x)^{3x} \left[ \ln (8x^3) + 3 \right] \qquad (x>0) \end{align*}

which coincides with the derivative obtained a bit earlier. Magic!

The Graph of (2x)^(3x) and Its Derivative

The graphs of $\displaystyle (2x)^{3x}$ and its derivative. By the way can you guess what that minimum is?

Example 6 — ${\left( \sqrt{x}^{\cos x}\right)}^{\ln x}$

So, what about the cases involving higher exponentiation like this one? Well, no sweat here, as this is not a real nested power function anyway. Indeed, using the Generalized Power Rule for Logarithm, one can collapse it back into an ordinary power function as follows:

\begin{align*} {\left( \sqrt{x}^{\cos x}\right)}^{\ln x} & = \sqrt{x}^{\cos x \ln x} \qquad (x>0)  \end{align*}

Once here, it can be seen that not only is $\displaystyle \sqrt{x}^{\, \cos x \ln x}$ well-defined on $\mathbb{R_+}$, but both the base function $\sqrt{x}$ and the exponent function $\cos x \ln x$ are differentiable on $\mathbb{R_+}$ as well. In which case, applying the Exponent Rule yields that:

\begin{align*} \left( \sqrt{x}^{\, \cos x \ln x} \right)’ & = \sqrt{x}^{\, \cos x \ln x} \left[ \left( -\sin x \ln x + \frac{\cos x}{x} \right) \ln \sqrt{x} + \frac{1}{2\sqrt{x}} \frac{\, \cos x \ln x}{\sqrt{x}} \right]  \\ & = \sqrt{x}^{\, \cos x \ln x} \left[ -\frac{1}{2}\sin x (\ln x)^2 + \frac{\cos x \ln x}{x} \right]  \qquad (x>0) \end{align*}

which gets a bit lengthy. But the point is that it is doable, so that if we put everything together, we get that:

\begin{align*} \left[ {\left( \sqrt{x}^{\, \cos x}\right)}^{\ln x} \right]’ & = \left( \sqrt{x}^{\, \cos x \ln x} \right)’ \\ & = {\left( \sqrt{x}^{\, \cos x}\right)}^{\ln x} \left[ -\frac{1}{2}\sin x (\ln x)^2 + \frac{\cos x \ln x}{x} \right] \qquad (x>0) \end{align*}

and here’s a picture of the graph as a bonus.

\left(\sqrt{x}^{\cos x}\right)^{\ln \ x}

Here’s the graph of the original function $\displaystyle {\left( \sqrt{x}^{\cos x}\right)}^{\ln x}$ (red) and its derivative (green) . Talk about some crazy oscillation!

Example 7 — $\sqrt{x}^{\cos x^{\ln x}}$

Now you might be asking, “isn’t that the one we’ve just worked on a bit earlier?” Actually, not quite. This is due to the not-so-secret convention that:

\begin{align*} \sqrt{x}^{\, \cos x^{\, \ln x}} & \stackrel{df}{=} \sqrt{x}^{\, (\cos x^{\, \ln x} ) } \end{align*}

In this case, the Power Rule for logarithm simply won’t work. But then, that’s not to say that it precludes us from calculating the derivative recursively either.

To see how, let’s begin by focusing on the exponent function $\displaystyle \cos x^{\, \ln x}$, from which we can see that:

  1. $\ln x$ is only defined when $x>0$.
  2. $\cos x>0$ only when $\displaystyle -\frac{\pi}{2} < x < \frac{\pi}{2} \pmod{2\pi}$.

So that if we let $I$ denote the set satisfying both restrictions. That is,

\begin{align*}I \stackrel{df}{=} \left\{ x \in \mathbb{R_+} \mathrel{\Big|} -\frac{\pi}{2} < x < \frac{\pi}{2} \,\, (\text{mod } 2\pi ) \right\} \end{align*}

then one can see that $\displaystyle \cos x^{\, \ln x}$ is well-defined on $I$, with both $\cos x$ and $\ln x$ being differentiable on $I$ as well. In which csae, the Exponent Rule kicks in, yielding that:

\begin{align*} \left( \cos x^{\, \ln x} \right)’ & = \cos x^{\, \ln x} \left[ \frac{1}{x} \ln (\cos x) + (-\sin x) \frac{\ln x}{\cos x} \right] \\ & = \cos x^{\, \ln x} \left[ \frac{\ln (\cos x)}{x} – \tan x \ln x  \right] \qquad (x \in I) \end{align*}

which takes care of the derivative of the exponent function. Now, if we just backtrack a bit to the original function, then it shouldn’t be hard to to see that:

  1. $\displaystyle \sqrt{x}^{\, (\cos x^{\, \ln x})}$ is well-defined on $I$.
  2. Both the base function $\sqrt{x}$ and the exponent function $\cos x^{\, \ln x}$ are differentiable on $I$.

which can only mean one thing — that $\displaystyle \sqrt{x}^{\, (\cos x^{\ln x})}$ is very well differentiable on $I$ as well. Here, applying the Exponent Rule a second time, we obtain that:

\begin{align*} \left[ \sqrt{x}^{\, ({\cos x}^{\, \ln x} )} \right]’ & = \sqrt{x}^{\, ({\cos x}^{\, \ln x} )} \left[ \left( {\cos x}^{\, \ln x}\right)’ \ln \sqrt{x} + \frac{1}{2\sqrt{x}} \frac{{\cos x}^{\, \ln x}}{\sqrt{x}} \right] \\ & = \cdots \end{align*}

Well, guess we’ll stop here and leave the algebra up to you. But hey, if you are proficient enough to have gone this far, you can finish that up right? 🙂

And for your curiosity, here’s the graph of the original function and its derivative for the record. Notice the stark contrast between this and the graphs from Example 6.

root x to (cos x to ln x)

The graph of the function $\displaystyle \sqrt{x}^{\, \cos x^{\, \ln x}}$ and its derivative. Pretty cool. Right?

Closing Words

Ouf! That was a bit of symbol-crunching up there! But hopefully it illustrates why the Exponent Rule can be a valuable asset in our arsenal of derivative rules. While for simple power function, this approach might seem like an overkill, for repeatedly-exponentiated power functions with one nested inside another, it becomes readily apparent that the Exponent Rule is absolutely the way to go.

In addition to automating the differentiation process for power functions, the Exponent Rule — especially when combined with other traditional derivative rules — can really do wonder in terms of tapping into functions that are previously too intimidating / tedious to tackle — such as the ones we have a hard time finding in a typical calculus textbook for instance. 🙂

In any case, let’s leave it at that and call it a day for now. By the way, here’s an interactive table summarizing what we have discovered thus far — for the sake of completeness:

Constant Rule
Power Rule
Sum Rule
Difference Rule
Product Rule
Reciprocal Rule
Quotient Rule
Chain Rule
Inverse-Function Rule

Given a base function $f$ and an exponent function $g$, if:

  1. The power function $f^g$ is well-defined on an interval $I$ (i.e.,$f$ and $g$ both well-defined on $I$, with $f>0$ on $I$)
  2. Both $f$ and $g$ are differentiable on $I$.

then the function $f^g$ is differentiable on $I$ as well. In addition:

\begin{align*} (f^g)’ = f^g \left( g’ \ln f + f’ \frac{g}{f} \right) \qquad (x \in I) \end{align*}

$\displaystyle (\pi^x)’ = \pi^x \ln \pi \quad (x \in \mathbb{R})$
$\displaystyle (b^x)’ = b^x \ln b \quad (x \in \mathbb{R})$
$\displaystyle (x^{\ln x})’ = x^{\ln x} \, \frac{2 \ln x}{x} \quad (x>0) $
$\displaystyle \left[ (x^2+1)^{\sin x} \right]’ = (x^2+1)^{\sin x} \left[ \cos x \ln (x^2+1) + \frac{2x \sin x}{x^2+1} \right] \quad (x \in \mathbb{R}) $
$\displaystyle \left[ (2x)^{3x}\right]’ = (2x)^{3x} \left[ 3 \ln(2x) + 3 \right] \quad (x>0) $
$\displaystyle \left[ {\left( \sqrt{x}^{\, \cos x}\right)}^{\ln x} \right]’ = {\left( \sqrt{x}^{\, \cos x}\right)}^{\ln x} \left[ -\frac{1}{2}\sin x (\ln x)^2 + \frac{\cos x \ln x}{x} \right] \quad (x>0) $
$\displaystyle \left[ \sqrt{x}^{\, ({\cos x}^{\, \ln x} )} \right]’ = $ you know what how

All right! Here marks the end of another relatively-short, 3000-words-ish  derivative module to you. Hey, interested in launching a petition calling for the inclusion of Exponent Rule into your textbook? Or maybe you want a petition precisely to prevent this absolute horror from ever materializing? Either way, why not let us know by dropping a comment below, or by voicing your appraisal/opposition on Facebook or Twitter!  🙂

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